I came across Probability Integral transform which states that if random variable $X$ has distribution function $F(x)$ then $F(X)$ has Uniform distribution on $[0,1]$.
I was applying this concept to the case of Exponential random variable.
whose PDF is $f(x) = \lambda e^{-\lambda x}$ for $x > 0$ and $f(x) = 0$ otherwise.
Whose CDF is $F(x) = 1 -e^{-\lambda x}$ for $x > 0$ and $0$ if $x \leq 0$.
Now $F(X) = 1 -e^{-\lambda X}$ for $X > 0$ and $0$ if $X \leq 0$.
Now what is the distribution of $F(X)$ , observing it as the function of the random variable $X$.
Applying the formula for $X $ has density function $f(x)$ then for $Y =g(X)$ the density function of $Y$ is $f(g^{-1}(y) |\frac{d(g^{-1}(y))}{dy}|) ; 0<y<1$ and $0$ otherwise.
Applying the above to our function of rv $X$ which is $F(X)$ we get –
the density function of $F(X)$ as $\frac{y}{\lambda (1-y)} ; o<y<1$ and $0$ otherwise.is this correct?
$\textbf{EDIT :}$
I got $g^{-1}(y) = \frac{-1}{\lambda}\ln(1-y)$ , $f(g^{-1}(y)) = y$ and $\frac{d(g^{-1}(y))}{dy} = \frac{1}{\lambda(1-y)}$.
So combining we get the density function of $F(X) $ say $h(y) = \frac{y}{\lambda(1-y)}$.
Now I am in doubt that I should have been getting Uniform density function but the above doesnot seem to follow Uniform distribution?
Best Answer
The proof of the fact for continuous distributions that therefore have invertible CDFs is: $$ P(F(X)\le z) = P(X \le F^{-1}(z)) = F((F^{-1}(z)) = z.$$ (The generalization to arbitrary distributions is just a matter of properly defining a generalized $F^{-1},$ but that's not important here.)
So you are making a calculation error.
Doing it the other way with the PDF, you have $Y = 1-\exp(-\lambda X)$ so that $X =-\frac{\ln(1-Y)}{\lambda}.$ So by the standard formula, $$f_Y(y) = \left|\frac{dx}{dy}\right|f_X\left(\frac{-\ln(1-y)}{\lambda}\right) = \frac{1}{\lambda(1-y)}\lambda e^{-\lambda \left(-\frac{\ln(1-y)}{\lambda}\right)}=1.$$
So either way, you get a uniform distribution, as promised.